Relative Complement inverts Subsets/Proof 2

Theorem

Let $S$ be a set.

Let $A \subseteq S, B \subseteq S$ be subsets of $S$.

Then:

$A \subseteq B \iff \relcomp S B \subseteq \relcomp S A$

where $\complement_S$ denotes the complement relative to $S$.

Proof

Sufficient Condition

Let $A \subseteq B$.

Then:

\(\ds x\)

\(\in\)

\(\ds \relcomp S B\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(\in\)

\(\ds S \setminus B\)

Definition of Relative Complement

\(\ds \leadsto \ \ \)

\(\ds x\)

\(\notin\)

\(\ds B\)

Definition of Set Difference

\(\ds \leadsto \ \ \)

\(\ds x\)

\(\notin\)

\(\ds A\)

Definition of Subset

\(\ds \leadsto \ \ \)

\(\ds x\)

\(\in\)

\(\ds S \setminus A\)

Definition of Set Difference

\(\ds \leadsto \ \ \)

\(\ds x\)

\(\in\)

\(\ds \relcomp S A\)

Definition of Relative Complement

\(\ds \leadsto \ \ \)

\(\ds \relcomp S B\)

\(\subseteq\)

\(\ds \relcomp S A\)

Definition of Subset

$\Box$

Necessary Condition

\(\ds \relcomp S B\)

\(\subseteq\)

\(\ds \relcomp S A\)

\(\ds \leadsto \ \ \)

\(\ds \relcomp S {\relcomp S A}\)

\(\subseteq\)

\(\ds \relcomp S {\relcomp S B}\)

from Sufficient Condition

\(\ds \leadsto \ \ \)

\(\ds A\)

\(\subseteq\)

\(\ds B\)

Relative Complement of Relative Complement

$\blacksquare$